Michael has a telecontrol robot. One day he put the robot on a loop with n cells. The cells are numbered from 1 to n clockwise. 



At first the robot is in cell 1. Then Michael uses a remote control to send m commands to the robot. A command will make the robot walk some distance. Unfortunately the direction part on the remote control is broken, so for every command the robot will chose a direction(clockwise or anticlockwise) randomly with equal possibility, and then walk w cells forward. 
Michael wants to know the possibility of the robot stopping in the cell that cell number >= l and <= r after m commands. 

Input

There are multiple test cases. 
Each test case contains several lines. 
The first line contains four integers: above mentioned n(1≤n≤200) ,m(0≤m≤1,000,000),l,r(1≤l≤r≤n). 
Then m lines follow, each representing a command. A command is a integer w(1≤w≤100) representing the cell length the robot will walk for this command.   
The input end with n=0,m=0,l=0,r=0. You should not process this test case. 

Output

For each test case in the input, you should output a line with the expected possibility. Output should be round to 4 digits after decimal points.

Sample Input

3 1 1 2
1
5 2 4 4
1
2
0 0 0 0

Sample Output

0.5000
0.2500

题意：给你一个n，m，l，r。n个格子围成一个圈，一个机器人从第一个格子开始，经过m个步，每个步知道走几格，但是不知道方向，问你最后在l到r之间的概率。

第一次写概率dp，时间上不是很理想，但是感觉最直观了，而且也过了，就不改了，多做些再想怎么优化吧

#include<cstdio>
using namespace std;
double dp[2][205];
int main(){
	int n,m,l,r;
	while(scanf("%d%d%d%d",&n,&m,&l,&r)){
		if(n==0&&m==0&&l==0&&r==0) break;
		for(int i=0;i<n;i++) dp[0][i]=0;
		dp[0][0]=1;
		int now=0;
		while(m--){
			int w;
			scanf("%d",&w);
			for(int i=0;i<n;i++)
			dp[now^1][i]=0.5*dp[now][(i-w+n)%n]+0.5*dp[now][(i+w)%n];
			now^=1;
		}
		double ans=0;
		for(int i=l-1;i<=r-1;i++)
		ans+=dp[now][i];
		printf("%.4lf\n",ans);
	}
	return 0;
}